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30 November, 12:42

A chemist obtains 500.0 mL of a solution containing an unknown concentration of calcium iodide, CaI2. He pipets 15 mL of this solution into a 100 mL volumetric flask and dilutes to the mark. He then pipettes 10 mL of this diluted solution into a 25 mL volumetric flask and dilutes to the mark. He analyzes some of the solution from the final volumetric flask and finds that the iodide ion concentration is 0.41 M. (Note: in solution, calcium iodide breaks apart into one Ca2 + ion for every two I - ions, so a solution that is 1.0 M in CaI2 is 2.0 M in I-.) Determine the molar concentration of calcium iodide in the original solution. LaTeX: CaI _{2} (s) → Ca^{ 2+} (aq) + 2 I^{-} (aq)

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  1. 30 November, 13:06
    0
    3.41735M

    Explanation:

    Step1: Balance the equation

    CaI2 (s) → Ca2 + (aq) + 2I - (aq)

    We have 15 mL of the original sample and pipet it in a 100mL solution.

    100/15 = 6.67 ⇒ this is a 1:6.67 dilution

    We take 10mL of this dilution into a 25 mL flask ⇒this is a 1:2.5 dilution

    6.67 * 2.5 = 16.675

    ⇒the total dilution is 1:16.675

    Let's imagine the conc of CaI2 [CaI2] = 1M

    ⇒ 15mL of this 1M conc in a 100mL = 0.15M

    ⇒10mL of this 0.15 M in 25mL = 0.06M

    1M original conc / O. O6M final concentration = 16.67

    ⇒this means the dilution is 16.67 (1:16.67)

    After analyzes we found that the iodide ion concentration = 0.41 M

    ⇒original conc / 0.41 M = 16.67

    original conc I - = 16.67 * 0.41M = 6.8347 M

    Since we have one Ca2 + ion for every 2 I - ions, we have to divide this concentration by 2 to find the [Ca2+]

    6.8347 / 2 = 3.41735 M = [Ca2+] = [CaI2]

    The concentration calcium iodide = 3.41735M
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